This question asks you to identify the incorrect thermodynamic expression among the given options. The core concept here is the relationship between Gibbs free energy (ΔG), the equilibrium constant (K), and other thermodynamic quantities like enthalpy (ΔH) and entropy (ΔS).

The fundamental equation linking these is:

$\mathrm{\Delta G}°=\mathrm{\Delta H}°-T\mathrm{\Delta S}°$

This equation defines the standard Gibbs free energy change.

Another crucial relationship connects the standard Gibbs free energy change to the equilibrium constant of a reaction:

$\mathrm{\Delta G}°=-RT \ln K$

This means the more negative ΔG° is, the larger the equilibrium constant K will be, favoring the products.

We can combine these two equations. Substituting the expression for ΔG° from the first equation into the second gives us:

$-RT \ln K=\mathrm{\Delta H}°-T\mathrm{\Delta S}°$

Now, let's solve this combined equation for lnK. We divide both sides by -RT:

$\ln K=-\frac{\mathrm{\Delta H}°-T\mathrm{\Delta S}°}{RT}$

This is the correct form of the equation. Let's compare it to the first option:

$\mathrm{lnK}=\frac{\mathrm{\Delta H}\mathrm{°}-\mathrm{T\Delta S}\mathrm{°}}{\mathrm{RT}}$

Notice the missing negative sign in the numerator. The first option is incorrect because it is missing the crucial negative sign. The correct equation is $\ln K=-\frac{\mathrm{\Delta H}°-T\mathrm{\Delta S}°}{RT}$.

The other options are correct:

• Option 2: The work done in a reversible isothermal expansion for an ideal gas is indeed $w_{\mathrm{rev}}=-nRT \ln \frac{V_f}{V_i}$.
• Option 3: For a spontaneous process, the total entropy change (ΔS_total) is positive. The relationship $\frac{\mathrm{\Delta G}}{\mathrm{\Delta S}\mathrm{total}}=-T$ is a valid thermodynamic identity linking the system's Gibbs free energy change to the total entropy change of the universe.
• Option 4: Starting from the correct relationship $\mathrm{\Delta G}°=-RT \ln K$, we can solve for K by dividing both sides by -RT and then exponentiating: $\ln K=-\frac{\mathrm{\Delta G}\mathrm{°}}{RT}$, therefore $K=e^{-\mathrm{\Delta G}\mathrm{°}/RT}$. This option is correct.

Related Concepts and Formulae

Gibbs Free Energy (G): A thermodynamic potential that predicts the direction of chemical reactions and determines the maximum amount of reversible work that can be performed by a system at constant temperature and pressure. A negative ΔG indicates a spontaneous process.

Key Formulae:

• Definition: $G=H-TS$
• Change at standard state: $\mathrm{\Delta G}°=\mathrm{\Delta H}°-T\mathrm{\Delta S}°$
• Relation to Equilibrium Constant: $\mathrm{\Delta G}°=-RT \ln K$
• Combined Form (van 't Hoff equation): $\ln K=-\frac{\mathrm{\Delta H}°}{RT}+\frac{\mathrm{\Delta S}°}{R}$